# BALANCING OF ROTATING ... effect of the centrifugal force of the first mass is called balancing of...

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Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 1.1

1 BALANCING OF ROTATING MASSES

Course Contents

1.1 Introduction

1.2 Static Balancing

1.3 Types of Balancing

1.4 Balancing of Several Masses

Rotating in the Same Plane

1.5 Dynamic Balancing

1.6 Balancing of Several Masses

Rotating in the different

Planes

1.7 Balancing Machines

1. Balancing of Rotating Masses Dynamics of Machinery (2161901)

Prepared By: Vimal Limbasiya Department of Mechanical Engineering Page 1.2 Darshan Institute of Engineering & Technology, Rajkot

1.1 Introduction

Often an unbalance of forces is produced in rotary or reciprocating machinery due

to the inertia forces associated with the moving masses. Balancing is the process of

designing or modifying machinery so that the unbalance is reduced to an

acceptable level and if possible is eliminated entirely.

Fig. 1.1

A particle or mass moving in a circular path experiences a centripetal acceleration

and a force is required to produce it. An equal and opposite force acting radially

outwards acts on the axis of rotation and is known as centrifugal force [Fig. 1.1(a)].

This is a disturbing force on the axis of rotation, the magnitude of which is constant

but the direction changes with the rotation of the mass.

In a revolving rotor, the centrifugal force remains balanced as long as the centre of

the mass of the rotor lies on the axis of the shaft. When the centre of mass does

not lie on the axis or there is an eccentricity, an unbalanced force is produced [Fig.

1.1(b)]. This type of unbalance is very common. For example, in steam turbine

rotors, engine crankshafts, rotary compressors and centrifugal pumps.

Most of the serious problems encountered in high-speed machinery are the direct

result of unbalanced forces. These forces exerted on the frame by the moving

machine members are time varying, impart vibratory motion to the frame and

produce noise. Also, there are human discomfort and detrimental effects on the

machine performance and the structural integrity of the machine foundation.

The most common approach to balancing is by redistributing the mass which may

be accomplished by addition or removal of mass from various machine members.

There are two basic types of unbalance-rotating unbalance and reciprocating

unbalance – which may occur separately or in combination.

1.2 Static Balancing:

A system of rotating masses is said to be in static balance if the combined mass

centre of the system lies on the axis of rotation.

Dynamics of Machinery (2161901) 1. Balancing of Rotating Masses

Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 1.3

1.3 Types of Balancing:

There are main two types of balancing conditions

(i) Balancing of rotating masses

(ii) Balancing of reciprocating masses

(i) Balancing of Rotating Masses

Whenever a certain mass is attached to a rotating shaft, it exerts some centrifugal

force, whose effect is to bend the shaft and to producevibrations in it. In order to prevent

the effect of centrifugal force, another mass is attached to the opposite side of theshaft,

at such a position so as to balance the effect of the centrifugal force of the first mass.

This is done in such away that the centrifugal forces of both the masses are made to be

equal and opposite. The process of providing the second mass in order to counteract the

effect of the centrifugal force of the first mass is called balancing of rotating masses.

The following cases are important from the subject point of view:

1. Balancing of a single rotating mass by a single mass rotating in the same plane.

2. Balancing of different masses rotating in the same plane.

3. Balancing of different masses rotating in different planes.

1.4 Balancing of Several Masses Rotating in the Same Plane

Consider any number of masses (say four) of magnitude m1, m2, m3 and m4 at

distances ofr1, r2, r3 and r4 from the axis of the rotating shaft. Let 1, 2,3and 4be

the angles of these masses with the horizontal line OX, as shown in Fig. 1.2 (a). Let

these masses rotate about an axis through O and perpendicular to the plane of

paper, with a constant angular velocity of rad/s.

(a)Space diagram. (b) Vector diagram.

Fig. 1.2 Balancing of several masses rotating in the same plane.

The magnitude and position of the balancing mass may be found out analytically

or graphically as discussed below:

1. Analytical method

1. Balancing of Rotating Masses Dynamics of Machinery (2161901)

Prepared By: Vimal Limbasiya Department of Mechanical Engineering Page 1.4 Darshan Institute of Engineering & Technology, Rajkot

Each mass produces a centrifugal force acting radially outwards from the axis of

rotation. Let F be the vector sum of these forces.

F = m1r1 2+ m2r2

2 +m3r3 2+ m4r4

2

The rotor is said to be statically balanced if the vector sum F is zero.

If F is not zero, i.e., the rotor is unbalanced, then produce a counterweight

(balance weight) of mass mc, at radius rc to balance the rotor so that

m1r1 2+ m2r2

2 +m3r3 2+ m4r4

2 + mcrc 2 = 0

m1r1+ m2r2 +m3r3+ m4r4 + mcrc = 0

The magnitude of either mc or rc may be selected and of other can be calculated.

In general, if mr is the vector sum of m1.r1, m2.r2, m3.r3, m4.r4, etc., then

mr +mcrc = 0

To solve these equation by mathematically, divide each force into its x and z

components, mrcos + mcrccosc = 0

and mrsin +mcrcsinc= 0

mcrccosc= −mrcos…………………………(i)

and mcrcsinc= −mrsin............................(ii)

Squaring and adding (i) and (ii),

mcrc = 𝑚𝑟𝑐𝑜𝑠 ² + 𝑚𝑟𝑠𝑖𝑛 ²

Dividing (ii) by (i),

𝑡𝑎𝑛𝑐 = −𝑚𝑟𝑠𝑖𝑛

−𝑚𝑟𝑐𝑜𝑠

The signs of the numerator and denominator of this function identify the quadrant

of the angle.

2. Graphical method

First of all, draw the space diagram with the positions of the several masses, as

shown in Fig. 1.2 (a).

Find out the centrifugal force (or product of the mass and radius of rotation)

exerted byeach mass on the rotating shaft.

Now draw the vector diagram with the obtained centrifugal forces (or the product

of themasses and their radii of rotation), such that ab represents the centrifugal

force exerted by the mass m1 (or m1.r1) in magnitude and direction to some

suitable scale. Similarly, draw bc, cd and de to represent centrifugal forces of

other masses m2, m3 and m4 (or m2.r2,m3.r3 and m4.r4).

Now, as per polygon law of forces, the closing side ae represents the resultant

force inmagnitude and direction, as shown in Fig. 1.2 (b).

The balancing force is, then, equal to resultant force, but in opposite direction.

Dynamics of Machinery (2161901) 1. Balancing of Rotating Masses

Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 1.5

Now find out the magnitude of the balancing mass (m) at a given radius of

rotation (r), such that

m.r.2= Resultant centrifugal force

or m.r = Resultant of m1.r1, m2.r2, m3.r3 and m4.r4

(In general for graphical solution, vectors m1.r1, m2.r2, m3.r3, m4.r4, etc., are added. If

they close in a loop, the system is balanced. Otherwise, the closing vector will be

giving mc.rc. Its direction identifies the angular position of the countermass relative

to the other mass.)

Example 1.1 :A circular disc mounted on a shaft carries three attached masses of 4 kg, 3 kg

and 2.5 kg at radial distances of 75 mm, 85 mm and 50 mm and at the angular positions of

45°, 135° and 240° respectively. The angular positions are measured counterclockwise from

the reference line along the x-axis. Determine the amount of the countermass at a radial

distance of 75 mm required for the static balance.

m1 = 4 kg r1 = 75 mm 1 = 45°

m2 = 3 kg r2= 85 mm 2 = 135°

m3 = 2.5 kg r3 = 50 mm 3 = 240°

m1r1 = 4 x 75 = 300 kg.mm

m2r2 = 3 x 85 = 255 kg.mm

m3r3 = 2.5 x 50 = 125 kg.mm

Analytical Method:

mr +mcrc = 0

300 cos45°+ 255 cos135° + 125 cos240° + mcrccosc= 0 and

300 sin 45°+ 255 sin 135° + 125 sin 240° + mcrcsinc= 0

Squaring, adding and then solving,

2

2

(300 cos 45 255 cos 135 125 cos240 )

(300 sin45 255 sin 135 125 sin240 ) C Cm r

2 275

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